Definition talk:Filtration of Sigma-Algebra

Regarding the question if $\bigcup \FF_n = \Sigma$ should be imposed in this definition: Williams does not seem to define it that way. He then defines $\mathcal F_\infty = \map \sigma {\bigcup_n \mathcal F_n} \subseteq \mathcal F$ which would be uninteresting with that condition. Caliburn (talk) 17:28, 6 May 2022 (UTC)

Hmm... I see. That's a bit confusing regarding the naming, as then $\Sigma$ is not well-defined by the filtration. Maybe other sources or further coverage will give us more clarity going forward. — Lord_Farin (talk) 17:53, 6 May 2022 (UTC)

Do you have a source on hand using this definition? Wikipedia also does not have the union condition and nor did my probability lecture course. I remember you saying something to the effect of that we're not ready for non-equivalent formulations, though. Caliburn (talk) 18:16, 6 May 2022 (UTC)

Just the fact that $\Sigma$ is mentioned. I guess all is fine, then. Forget I said anything. — Lord_Farin (talk) 18:32, 6 May 2022 (UTC)

For example, suppose that $X$ is a topological space and $\Sigma = \map {\BB} X$ is its Borel algebra. When we consider a specific filtration $\sequence {\FF_n}_{n \mathop \in \N}$, then $\mathcal F_\infty$ is typically not $\Sigma$. --Usagiop (talk) 23:36, 28 October 2022 (UTC)